# Check for convergence or divergence in the following sequences?

## just tell me which test to use Currently both come up inconclusive ($L = 1$) with the Root Test ? ${a}_{n} = {\left(1 + \frac{3}{n}\right)}^{4 n}$ ${a}_{n} = {\left(\frac{n}{n + 3}\right)}^{n}$

Apr 24, 2018

A) Converges to ${e}^{12}$ B) Converges to ${e}^{-} 3$

#### Explanation:

The Ratio and Root Tests are used for determining the behavior of infinite series rather than infinite sequences. Here, they won't really be of use.

Checking the convergence or divergence of a sequence is much simpler, and only requires taking the limit to infinity of the sequence.

If ${\lim}_{n \to \infty} {a}_{n} = L \ne \pm \infty$ then ${a}_{n}$ is said to converge to the finite value $L .$

A) Here, if we take the limit, we see

${\lim}_{n \to \infty} {\left(1 + \frac{3}{n}\right)}^{4 n} = {\left(1 + \frac{3}{\infty}\right)}^{4 \cdot \infty} = {1}^{\infty}$, an indeterminate form.

As a result, we're going to want to use l'Hospital's Rule.

But sequences are not differentiable, so we'll rewrite an equation $y$ in terms of $x$ and set up for l'Hospital's Rule as follows:

$y = {\left(1 + \frac{3}{x}\right)}^{4 x}$

$\ln y = \ln \left[{\left(1 + \frac{3}{x}\right)}^{4 x}\right]$

Recall the logarithm exponent property, which tells us that $\ln \left({a}^{b}\right) = b \ln a :$

$\ln y = 4 x \ln \left(1 + \frac{3}{x}\right)$

Simplify the argument of the logarithm:

$1 + \frac{3}{x} = \frac{x + 3}{x}$

$\ln y = 4 x \ln \left(\frac{x + 3}{x}\right)$

Recall the logarithm quotient property, which tells us that $\ln \left(\frac{a}{b}\right) = \ln a - \ln b :$

$\ln y = 4 x \left(\ln \left(x + 3\right) - \ln x\right)$

Finally, we want this in rational function form for using l'Hospital's Rule. Multiplying by $4 x$ is equal to dividing by $\frac{1}{4 x}$:

$\ln y = \frac{\ln \left(x + 3\right) - \ln x}{\frac{1}{4 x}}$

Take the limit to infinity:

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \frac{\ln \left(x + 3\right) - \ln x}{\frac{1}{4 x}} = \frac{\infty - \infty}{0}$

This is definitely indeterminate, so differentiate the numerator and denominator:

=lim_(x->oo)(1/(x+3)-1/x)/(-1/(4x^2)

$\frac{1}{x + 3} - \frac{1}{x} = \frac{\cancel{x} - \left(\cancel{x} + 3\right)}{x \left(x + 3\right)} = - \frac{3}{x \left(x + 3\right)}$

So, we get

=lim_(x->oo)(cancel-3/(x(x+3)))/(cancel-1/(4x^2))=lim_(x->oo)3/((cancelx(x+3))/(4x^cancel2)

$= {\lim}_{x \to \infty} \frac{3}{\frac{x + 3}{4 x}} = \frac{3}{\frac{1}{4}} = 12$

Now that we know ${\lim}_{n \to \infty} \ln y ,$ we can find ${\lim}_{n \to \infty} y :$

lim_(n->oo)y=lim_(n->oo)e^lny=e^(lim_(n->oo)lny=e^12

The sequence converges to ${e}^{12.}$

B) This is also indeterminate: ${\lim}_{n \to \infty} {\left(\frac{n}{n + 3}\right)}^{n} = {1}^{\infty}$

So, proceed as we did in the previous problem:

$y = {\left(\frac{x}{x + 3}\right)}^{x}$

$\ln y = \ln \left[{\left(\frac{x}{x + 3}\right)}^{x}\right]$

$\ln y = x \ln \left(\frac{x}{x + 3}\right)$

$\ln y = x \left(\ln x - \ln \left(x + 3\right)\right)$

$\ln y = \frac{\ln x - \ln \left(x + 3\right)}{\frac{1}{x}}$

${\lim}_{x \to \infty} \ln y = \frac{\infty - \infty}{0}$ Indeterminate, differentiate numerator and denominator:

$= {\lim}_{x \to \infty} \frac{\frac{1}{x} - \frac{1}{x + 3}}{- \frac{1}{x} ^ 2}$

$\frac{1}{x} - \frac{1}{x + 3} = \frac{\cancel{x} + 3 - \cancel{x}}{x \left(x + 3\right)}$

$= {\lim}_{x \to \infty} \frac{- \frac{3}{x \left(x + 3\right)}}{\frac{1}{x} ^ 2}$
=lim_(x->oo)-3/((cancelx(x+3))/(x^cancel2)
$= {\lim}_{x \to \infty} - \frac{3}{\frac{x + 3}{x}} = {\lim}_{x \to \infty} - \frac{3}{1 + \frac{3}{x}} = - 3$

So ${\lim}_{x \to \infty} \ln y = - 3$ and

${\lim}_{x \to \infty} y = {\lim}_{x \to \infty} {e}^{\ln} y = {e}^{{\lim}_{x \to \infty} \ln y} = {e}^{-} 3$

The sequence converges to ${e}^{-} 3.$